【HDU - 4217 】Data Structure? （线段树求第k小数）

题干：

Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.
Original, there are N numbers, namely 1, 2, 3...N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.

Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1

Output

For each test case, output the case number first, then the sum.

Sample Input

Sample Output

Case 1: 3
Case 2: 14

解题报告：

这题是暑假集训的时候的线段树的题，用线段树维护在某一个区间还剩下几个数。跟这题对比【HDU - 4006】The kth great number

AC代码：

//又忘了开longlong了吧。。。
#include<bits/stdc++.h>using namespace std;
const int MAX =262144 + 100000;
int n,m;struct TREE {int l,r;int val;
} tree[MAX*4];//数组要不要再大一点？void pushup(int cur) {tree[cur].val = tree[cur*2].val+ tree[cur*2+1].val;
}
void build(int l,int r,int cur) {tree[cur].l = l;tree[cur].r = r;if(tree[cur].l == tree[cur].r) {tree[cur].val = 1;return ;}int m = (l + r)/ 2;build(l,m,cur*2);build(m+1,r,cur*2+1);pushup(cur);}
int query(int k,int cur) {if(tree[cur].l == tree[cur].r){//       tree[cur].val = 0;return tree[cur].l;}if(k > tree[cur*2].val) return query(k-tree[cur*2].val,cur*2+1);else return query(k,cur*2);
}
void update(int tar,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val = 0;return ;}if(tar <= tree[cur*2].r) update(tar,2*cur);else update(tar,2*cur+1);pushup(cur);
}int main()
{int t,tmp;int k,iCase = 0;;long long ans = 0;cin>>t;while(t--) {//初始化ans = 0;scanf("%d%d",&n,&m);build(1,n,1);while(m--) {scanf("%d",&k);tmp = query(k,1);update(tmp,1);ans += tmp;}printf("Case %d: %lld\n",++iCase,ans);}return 0 ;
}

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